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How can I know which resistor to chose? Answered

Is there any math calc to know which resistor will low the voltage to a desired voltage?

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As far as I know, one resistor on its own won't do that. You would need to resistors to form a voltage divider and then you would use the formulae in the link I already provided. The point between the two resistors should come to 3.7V. Or better, you can do as alex suggested and get a variable-controlled voltage regulator. You should be able to pick one up at Radioshack.

Using "voltage dropping resistors" was a common way to "regulate" voltage before voltage regulators were common. But it's a very inefficient way to drop voltage. They work because the device you're powering is itself a "load" on the circuit, and draws current. Since that inserted resistance is fixed, varying the current draw has to change the voltage, too (it's an "ohms law" thing.) Unless you know the approx load of the device itself, you cannot know the voltage drop (even if you know the value of the resistor.)

So, there's no such thing as a math procedure to know approximately how to drop from a determined voltage to another right? With load you are refering to voltage or energy loaded in the battery? Sorry, I'm new in electronics and I'm not an english native. Thanks

Yes, but only if you know the "load."

In it's most simple form, the load is equivalent to another resistor. It's more complicated, but we'll pretend your device (which we know nothing about) creates a "resistive load."

To find the approx load, you'll need an ammeter. Most cheap VOMs (Volt-ohm meter) are ammeters, too. You can find VOMs for less than \$5 USD.

To drop 12V to 3V:

With the ammeter connected in series between the 3V and the device you are powering, take a reading. You are measuring the current (which is amps--A, or 1/1000 of amps, which is milliAmps--mA.)

For example, let's say the device draws 15mA (0.015 A) at 3V.

Using Ohm's Law: R = E / I , or resistance = voltage / current

The "resistive load" of the device is 200 ohms ( 200 = 3.0 / 0.015)

You can low plug that value into the voltage divider calculation randolfo referenced (see the pic below of a divider.)

To solve for R1 (the unknown "voltage dropping resistor")

R1 = ((R2 * Vin) / Vout) - R2

((200 * 12) / 3) - 200 = 600 ohms for R1.

I.E., the voltage between R1 and R2 should be 3V.

This only works if the load (the device, R2) is resistive, and is at best an approximation if it's not.

Current draw (amps, milliamps) MUST be measured at the rated voltage. If the device runs on 3V normally, it must be read at 3V.